Let $f(x)=-2x^4-8x^3+180x^2$. For what values of $x$ does the graph of $f$ have a point of inflection? Choose all answers that apply: Choose all answers that apply: (Choice A) A $x=-5$ (Choice B) B $x=3$ (Choice C) C $x=5$ (Choice D) D $f$ has no points of inflection.
Explanation: We can find the inflection points of the graph of $f$ by looking for the intervals where its second derivative $f''$ is positive/negative. This analysis is very similar to finding minimum/maximum points, only instead of analyzing $f'$, we are analyzing $f''$. The second derivative of $f$ is $f''(x)=-24(x+5)(x-3)$. $f''(x)=0$ for $x=-5,3$. Since $f''$ is a polynomial, it's defined for all real numbers. Therefore, our possible inflection points are $x=-5$ and $x=3$. Our possible inflection points divide the number line into three intervals: $\llap{-}7$ $\llap{-}6$ $\llap{-}5$ $\llap{-}4$ $\llap{-}3$ $\llap{-}2$ $\llap{-}1$ $0$ $1$ $2$ $3$ $4$ $(-\infty, \llap{-}5)$ $( \llap{-}5,3)$ $(3,\infty)$ Let's evaluate $f''$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $f''(x)$ Verdict $(-\infty,-5)$ $x=-6$ $f''(-6)=-216<0$ $f$ is concave down $\cap$ $(-5,3)$ $x=0$ $f''(0)=360>0$ $f$ is concave up $\cup$ $(3,\infty)$ $x=4$ $f''(4)=-216<0$ $f$ is concave down $\cap$ We can see that the graph of $f$ changes concavity at both $x=-5$ and $x=3$. In conclusion, these are the values of $x$ where the graph of $f$ has a point of inflection: $x=-5$ $x=3$